This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Muraj, Hamza Homework 27 Due: Apr 5 2006, 4:00 am Inst: Florin 1 This printout should have 9 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 3) 0 points A uniform rod, supported and pivoted at its midpoint, but initially at rest, has a mass of 84 g and a length 5 cm. A piece of clay with mass 34 g and velocity 1 . 7 m / s hits the very top of the rod, gets stuck and causes the clay rod system to spin about the pivot point O at the center of the rod in a horizontal plane. Viewed from above the scheme is 1 . 7 m / s O 34 g 84 g (a) (b) (c) 5cm Figure: The piece of clay and rod: (a) before they collide, (b) at the time of the collision, and (c) after they collide. After the collisions the clayrod system has an angular velocity about the pivot. With respect to the pivot point O , what is the magnitude of the initial angular momen tum L i of the clayrod system? Correct answer: 0 . 1445 kg m 2 / s. Explanation: Let : m putty = m = 34 g = 0 . 034 kg , m rod = 2 m = 84 g = 0 . 084 kg , = 5 cm = 0 . 05 m , and v = 1 . 7 m / s . Basic Concepts: Conservation of angular momentum ~ L ~ L = ~r ~p = m~r ~v X ~ ext = d ~ L dt L z = I . Therefore, if the net external torque acting on a system is zero, the total angular momentum of that system is constant. Since the total external torque acting on the clayrod system is zero, the total angular momentum is a constant of motion. The total initial angular momentum L i is simply the angular momentum of the clay, since the rod is at rest initially L i = k ~r ~p k = m r v = m v 2 (1) = (0 . 034 kg)(1 . 7 m / s)(0 . 05 m) 2 = . 1445 kg m 2 / s . 002 (part 2 of 3) 0 points With respect to the pivot point O , what is the final moment of inertia I f of the clayrod system? Correct answer: 8 . 75 10 5 kg m 2 . Explanation: The final moment of inertia I f of the clay rod system is the moment of inertia of the rod plus the moment of inertia of the clay I f = I rod + I clay = 1 12 2 m 2 + m 2 2 = 5 12 m 2 (2) = 5 12 (0 . 034 kg)(0 . 05 m) 2 = 8 . 75 10 5 kg m 2 ....
View
Full
Document
This note was uploaded on 04/12/2010 for the course PHY 58195 taught by Professor Turner during the Spring '09 term at University of Texas at Austin.
 Spring '09
 Turner

Click to edit the document details